Solve ∫ (from 0 to 1) of 2x+3x^2 wrt x

$$\int _ { 0 } ^ { 1 } 2 x + 3 x ^ { 2 } d x$$

Evaluate

$$2$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2x+3x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x\mathrm{d}x+\int 3x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x\mathrm{d}x+3\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 2 times \frac{x^{2}}{2}.

$$x^{2}+3\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 3 times \frac{x^{3}}{3}.

$$x^{2}+x^{3}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$1^{2}+1^{3}-\left(0^{2}+0^{3}\right)$$

Simplify.

$$2$$